1Z0-811 Primitives, Strings and Operators Practice Question
A developer writes: char c = 'A'; int i = c + 1; System.out.println(i); What is the output?
⚠ Common exam trap
Oracle often tests the misconception that `char` and `int` cannot be added, or that the result remains a `char` and would print as a character, causing candidates to choose 'B' instead of the numeric value.
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
66
In Java, when a `char` is involved in arithmetic with an `int`, the `char` is promoted to its Unicode/ASCII numeric value. 'A' has the ASCII value 65, so `c + 1` becomes 66. The result is an `int`, and `System.out.println(i)` prints the integer 66.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✓
66
Why this is correct
'A' is 65, plus 1 equals 66.
- ✗
B
Why it's wrong here
Output is int, not char.
- ✗
Compilation error: cannot add char and int.
Why it's wrong here
Implicit promotion allowed.
- ✗
'A1'
Why it's wrong here
String concatenation not used.
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